HDU 2602 Bone Collector "01 Backpack template"

Bone Collector

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 40404 Accepted Submission (s): 16786

Problem Descriptionmany years ago, in Teddy ' s hometown there is a man who was called "Bone Collector". Collect varies of bones, such as dog ' s, cow ' s, also he went to the grave ...
The bone collector had a big bag with a volume of V, and along he trip of collecting there is a lot of bones, obviously , different bone have different value and different volume, now given the each bone's value along his trips, can you CALCU Late out the maximum of the total value the bone collector can get?


Inputthe first line contain a integer T, the number of cases.
Followed by T cases, each case three lines, the first line contain both integer n, V, (N <=, v <=) repr Esenting the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume for each bone.
Outputone integer per line representing the maximum of the total value (this number would be is less than 231).
Sample Input

15 101 2 3 4 55 4 3 2 1

Sample Output

14

Authorteddy
Sourcehdu 1st "Vegetable-birds Cup" programming Open Contest Code:

HDU 2602 Bone collector#include <stdio.h> #include <string.h> #include <algorithm> #define MAX (A, B) ( A>B?A:B) using namespace std;struct inin{int val;//value! int cos;//Volume! }boy[10001];int bag[10001];//the size of the table below a number of items total value! int main () {int T;int n,v;//n represents the number of backpacks, V represents the overall product of the backpack! scanf ("%d", &t), while (t--) {memset (boy,0,sizeof (Boy)),//Save the value and volume of an item! memset (bag,0,sizeof (bag));//Save the total value of I items! scanf ("%d%d", &n,&v),//n represents the number of backpacks, V represents the overall product of the backpack! for (int i=1;i<=n;i++) {scanf ("%d", &boy[i].val);//value}for (int j=1;j<=n;j++) {scanf ("%d", &boy[j].cos); /volume}for (int i=1;i<=n;i++) {for (int j=v;j>=boy[i].cos;j--)//The maximum value {Bag[j]=max (Bag[j],bag[j-boy[i]) that the backpack can load from the back forward. Cos]+boy[i].val);//See Put and do not put the article I item of his value which large, choose the value of larger put into the array bag inside}//meaning is not put on the basis of the largest volume of the present and not put, It is only on the basis of your current volume minus the volume of the backpack you are going to put on the largest amount of value that can be accommodated! }printf ("%d\n", Bag[v]);//Finally, after all the items have been traversed, the last greatest value is the request! }return 0;}

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HDU 2602 Bone Collector "01 Backpack template"